Problem 1069

Let $P$ be the foot of the perpendicular dropped from point $D$ to line $A C$. Line $A C$ is perpendicular to two intersecting lines $A C$ and $D D 1$ in the plane $D D 1 P$, so line $A C$ is perpendicular to this plane. Therefore, any line passing through point $P$ and perpendicular to $A C$ (or coinciding with it, line $K L$) lies in the plane $D D 1 P$. It is known that the lateral edge of a regular triangular pyramid is perpendicular to the intersecting lateral edge. Additionally, if a line $l$ and a plane $\alpha$ are perpendicular to the same line, then line $l$ either lies in plane $\alpha$ or is parallel to it. The intersecting lines $K L$ and $M N$ are perpendicular, and plane $D D 1 P$ is perpendicular to line $K L$, so line $M N$ either lies in plane $D D 1 P$ or is parallel to it. The second case is excluded because, by the problem's condition, point $N$ lies on line $D D 1$, i.e., it is a common point of line $D D 1$ and plane $D D 1 P$. Therefore, line $M N$ lies in plane $D D 1 P$. At the same time, point $M$ lies in plane $D D 1 C 1 C$ because it lies on line $E F$ of this plane. Therefore, point $M$ lies on line $D D 1$, the intersection of planes $D D 1 C 1 C$ and $D D 1 P$. Then $D P$ is the common perpendicular of the intersecting lines $D D 1$ and $K L$, and since the common perpendicular of opposite edges of a regular triangular pyramid passes through the midpoint of the base edge, $D$ is the midpoint of $M N$. Let $A B=4 a, B C=3 a$, $K L=2 b, M N=3 b$. Let $H$ be the center of the base $M N L$ of the regular pyramid $K L M N$. From the right triangle $K(H L$ we find that

$$\cos \angle K L H=\frac{H L}{K L}=\frac{2 . D L}{K L}=\frac{2}{3} \cdot \frac{M N \sqrt{3}}{2} K L=\frac{2}{3} \cdot \frac{3 b \sqrt{3}}{2} 2 b=\frac{\sqrt{3}}{2}$$

Therefore, $\angle D L P=\angle K L H=30^{\circ}$ and

$$D P=\frac{1}{2} D L=\frac{1}{2} \cdot \frac{3 b \sqrt{3}}{2}=\frac{3 b \sqrt{3}}{4}, K H=\frac{1}{2} K L=b$$

Let lines $E F$ and $C D$ intersect at point $G$. From the equality of triangles $M F D 1, E F C 1$ and $E G C$ it follows that $C G=F C 1=F D 1$, and from the similarity of triangles $M F D 1$ and $M G D-$

$$\frac{M D_{1}}{M D}=\frac{F D_{1}}{D G}=\frac{F^{\prime} D}{C D+C G}=\frac{2 a}{4 a+2 a}=\frac{1}{3}$$

Therefore,

$$D D 1=\frac{2}{3} M D=\frac{2}{3} \cdot \frac{1}{2} M N=\frac{1}{3} \cdot 3 b=b$$

From the right triangle $A C D$ we find that

$$A C=\sqrt{A D^{2}+C D^{2}}=\sqrt{9 a^{2}+16 a^{2}}=5 a, D P=\frac{A D C D}{A C}=\frac{3 a+4 a}{5 a}=\frac{12}{5} a$$

!
Then

$$\begin{aligned} V 1=S_{A B C D} \cdot D D 1 & =3 a \cdot 4 a \cdot b=12 a 2 b=12 a 2 b=12\left(\frac{5 b \sqrt{3}}{16}\right) 2 \cdot b=\frac{225 b^{2}}{64} \\ V 2 & =\frac{1}{3} S_{M N L} \cdot K H=\frac{1}{3} \cdot \frac{(3 b)^{2} \sqrt{3}}{4} \cdot b=\frac{3 b b^{3} \sqrt{3}}{4} \end{aligned}$$

Therefore,

$$\frac{V_{1}}{V_{2}}=\frac{\frac{22 \sqrt[2]{2} 5^{2}}{1}}{\frac{\sqrt{3} \sqrt{5} \sqrt{3}}{1}}=\frac{25 \sqrt{3}}{16}$$

## Answer

$\frac{25 \sqrt{3}}{16}$

16.